Question

A centripetal-acceleration addict rides in a uniform circular motion with period $\mathit{T}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$and radius r = 3.00 m. At one instant his acceleration is$\overrightarrow{\mathbf{a}}\mathbf{=}\left(6.00{\text{m/s}}^{\text{2}}\right)\hat{\mathbf{i}}\mathbf{+}\mathbf{}\left(-4.00{\text{m/s}}^{\text{2}}\right)\hat{\mathbf{j}}$. (a) At that instant, what are the values of$\overrightarrow{\mathbf{v}}\mathbf{·}\overrightarrow{\mathbf{a}}$? (b) At that instant, what are the values of$\overrightarrow{\mathbf{r}}\mathbf{}\mathbf{\times }\overrightarrow{\mathbf{a}}$?

Short answer

a) At the given instant the value $\overrightarrow{v}.\overrightarrow{a}$of is zero.

b) At the given instant the value $\overrightarrow{r}\times \overrightarrow{a}$of is zero.

Step-by-step solution

Given data

The radius is $\mathrm{r}=3.00\mathrm{m}$and acceleration $\overrightarrow{a}=(6.0\mathrm{m}/{\mathrm{s}}^2)\hat{i}+(-4\mathrm{m}/{\mathrm{s}}^2)\hat{j}$.

Understanding the concept

From the given situation, we can find the value of dot and cross product. The time period of motion by using the given frequency and from the given radius, it is easy to find the magnitude of acceleration.

During constant-speed circular motion, the velocity vector is perpendicular to the acceleration vector at every instant.

The acceleration in vector, at every instant, points towards the center of the circle whereas the position vector points from the center of the circle to the object in motion.

Formulae:

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{b}=ab.\cos \theta \left(1\right) \\ \overrightarrow{a}\times \overrightarrow{b}=ab.\sin \theta \left(2\right) \end{gathered}$$

(a) Calculate v→.a→

As velocity vector is perpendicular to the acceleration vector at every instant during constant speed circular motion. The angle between them is $90°$. As cosine of 90 is zero. Therefore, using equation (i), it can be stated that $\overrightarrow{v}.\overrightarrow{a}=0$.

(b) Calculate r→×a→

The angle between the position vector and acceleration vector is $180°$. As sin of $180°$is zero. Therefore, using equation (ii), it can be stated that, $\overrightarrow{r}\times \overrightarrow{a}=0$.